Is #y=x^2+1#, #y=2^(x^2)# a solvable system?

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Answer 1 · 2017-05-23T12:52:19

Yes. It is a solvable system.

Making #x^2=y-1# and substituting

#y = 2^(y-1)#

A natural solution is #y=1 rArr x= 0# and after a close view also #y=2 rArr x = pm 1#